Question: Graph this system of equations and solve. $y = x + 3$ $-4x-3y = 12$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
The y-intercept for the first equation is $3$ , so the first line must pass through the point $(0, 3)$ The slope for the first equation is $1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $1$ position to the right. $1$ position to the right. $1$ position up from $(0, 3)$ is $(1, 4)$ Graph the blue line so it passes through $(0, 3)$ and $(1, 4)$ Convert the second equation, $-4x-3y = 12$ , to slope-intercept form. $y = -\dfrac{4}{3} x - 4$ The y-intercept for the second equation is $-4$ , so the second line must pass through the point $(0, -4)$ The slope for the second equation is $-\dfrac{4}{3}$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move down (because it's negative) $3$ positions to the right. $4$ positions down from $(0, -4)$ is $(3, -8)$ Graph the green line so it passes through $(0, -4)$ and $(3, -8)$ The solution is the point where the two lines intersect. The lines intersect at $(-3, 0)$.